Poj2976 -- Dropping tests (0-1 score Planning)

Poj2976 -- Dropping tests (0-1 score Planning)

Poj2976: Question Link

N pairs are given. a maximum of k pairs can be excluded from them, and the maximum value of a/B is obtained.

0-1 score planning: x = sigma a/SIGMA B --> 0 = sigma a-x * Sigma B --> g (x) = max (sigma a-x * Sigma B ), we can find that g (x) is a monotonic decreasing function about x. When g (x) is 0, x is the maximum value. Because of this monotonicity, binary solutions can be used.

Assume that s is the required value.

G (x) = 0 --> x = s;

G (x)> 0 --> x <s;

G (x) <0 --> x> s;

For this question, g (x) = max (Σ (100 * ai-x * bi), because up to k can be discarded, so the negative values of the smallest k are discarded, you can get the maximum value.

I originally wanted to learn the largest density subgraph. I read the thesis and started from scratch.

 

#include 
 
  #include 
  
   #include 
   
    #include using namespace std ;#define eqs 1e-9struct node{    double a , b ;}p[1100] ;int n , k ;double c[1100] ;double solve(double s) {    double ans = 0 ;    for(int i = 0 ; i < n ; i++)        c[i] = 100.0*p[i].a - s*p[i].b ;    sort(c,c+n) ;    for(int i = 0 ; i < n ; i++) {        if( i < k ){            if( c[i] >= eqs ) ans += c[i] ;        }        else ans += c[i] ;    }    return ans ;}int main() {    int i , j ;    double low , mid , high , temp ;    while( scanf("%d %d", &n, &k) && n+k > 0 ) {        for(i = 0 , high = 0 ; i < n ; i++) {            scanf("%lf", &p[i].a) ;            high += p[i].a ;        }        for(i = 0 ; i < n ; i++)            scanf("%lf", &p[i].b) ;        low = mid = 0 ;        high *= 100.0 ;        while( high-low >= eqs ) {            mid = (high + low) / 2.0 ;            temp = solve(mid) ;            if( fabs(temp) < eqs ) break ;            else if( temp < 0 )                high = mid ;            else                low = mid ;        }        printf("%d\n", (int)(mid+0.5)) ;    }    return 0 ;}